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I'm new here on Mathematics and have only basic algebraic knowledge. I have a problem in how to solve the following equation: $$ P^x = R_0^x + R_1^x + ... +R_n^x $$ I know the value of P and the values of all $R$'s. Sometimes there are only two $R$'s but can also be $3, 4$ or $5$ $R$'s.

How do I solve for $x$? Is this possible trough algebra?

Any help would be greatly appreciated.

JH6
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    In general no, it is not possible to manipulate the symbols to get x = . you will need to use a numerical method like the Bisection Method or Newton's Method. – Paul Jan 04 '16 at 10:32
  • Thank you for your answer. I will look into the bisection and newton's method – JH6 Jan 04 '16 at 10:37
  • you could plot graphs of $y = P^x$ and $y=R_0^x+...R_n^x$ to start with. Where they cross is the x value you want. This will let you know you are on the right track with your solution and give you good starting values for a numerical method. – Paul Jan 04 '16 at 10:41
  • Can you tell us anything about the relative values of $P$ and the $R$'s ? –  Jan 04 '16 at 11:09

2 Answers2

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Setting $P^x=t$, the equation turns to

$$t=t^{\alpha_0}+t^{\alpha_1}+\cdots t^{\alpha_n}$$ where

$$\alpha_k=\frac{\log(R_k)}{\log(P)}.$$

Such "generalized polynomial" equations are known to have no solutions in radicals for rational exponents, in general.

Even with two $R$'s there is no analytical solution (except for very specific values of the $\alpha$'s). In practice, you will have to resort to numerical methods.

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Use a numerical method. I'd suggest using something like the secant or Brent's method, as they don't require derivatives (they get messy with such an expression, and costly to compute). Starting points could be taking, with the $R_i$ in increasing order:

$\begin{align} P^x &= R_1^x + \dotsb + R_n^x \\ 1 &= \left(\frac{R_1}{P}\right)^x + \dotsb + \left(\frac{R_n}{P}\right)^x \end{align}$

Rough bounds should be the solutions of:

$\begin{align} 1 &= n \left(\frac{R_1}{P}\right)^{x_a} \\ 1 &= n \left(\frac{R_n}{P}\right)^{x_b} \end{align}$

Or just ask your tame computer algebra system to solve it numerically for you.

vonbrand
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